The lifespans of lions in a particular zoo are normally distributed. The average lion lives $13.1$ years; the standard deviation is $2.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living less than $15.5$ years.
Answer: $13.1$ $10.7$ $15.5$ $8.3$ $17.9$ $5.9$ $20.3$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $13.1$ years. We know the standard deviation is $2.4$ years, so one standard deviation below the mean is $10.7$ years and one standard deviation above the mean is $15.5$ years. Two standard deviations below the mean is $8.3$ years and two standard deviations above the mean is $17.9$ years. Three standard deviations below the mean is $5.9$ years and three standard deviations above the mean is $20.3$ years. We are interested in the probability of a lion living less than $15.5$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the lions will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the lions will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $10.7$ years and the other half $({16\%})$ will live longer than $15.5$ years. The probability of a particular lion living less than $15.5$ years is ${68\%} + {16\%}$, or $84\%$.